Simple Integral, Still Having Trouble

1. Oct 2, 2011

Blues_MTA

1. The problem statement, all variables and given/known data
I am trying to take the integral of 1/((-4t2+4t+3)1/2)

I know that i need to complete the square and it should come out to an inverse sin function but i dont understand how the completed square in the denominator is equal to (4-(2t-1)2)1/2

2. Relevant equations

3. The attempt at a solution

2. Oct 2, 2011

issacnewton

just expand the square and you see they are equal

3. Oct 2, 2011

Blues_MTA

But how do you come to that? I understand its equal, i just dont know how they calculated it, am I missing something completely obvious or...?

4. Oct 2, 2011

issacnewton

have you studied how to complete square ? if you have a quadratic equation

$$ax^2+bx+c = 0$$ then you add and subtract middle term square divided by

4 times the first term

$$ax^2+bx+c+\frac{b^2x^2}{4ax^2}-\frac{b^2x^2}{4ax^2}+c=0$$

$$a(x^2+\frac{b}{a}x+\frac{b^2}{4a^2})=\frac{b^2}{4a}-c$$

$$a(x+\frac{b}{2a})^2=\frac{b^2}{4a}-c$$

so use this procedure. in your case there is just a quadratic term, no equation. but
this approach can still be used

5. Oct 2, 2011

Blues_MTA

Yeah, I have, and for some reason this problem completely stumps me, If you use those formulas , you end up with 4(t+1/2)^2 = -2

I dont see how that is equivalent to 4-(2t-1)^2

6. Oct 2, 2011

issacnewton

$$-4t^2+4t+3$$

$$-4t^2+4t +\frac{(4t)^2}{4(-4t^2)}-\frac{(4t)^2}{4(-4t^2)}+3$$

$$-4t^2+4t-1+1+3$$

$$-(4t^2-4t+1)+4$$

$$4-(4t^2-4t+1)$$

$$4-(2t-1)^2$$

$$\smile$$

7. Oct 2, 2011

Blues_MTA

Ok, never mind i see that they are equivalent but how do you know to rearrange them in that manner to take the integral? is there any sort of method or is it just through practice

8. Oct 2, 2011

Blues_MTA

oooook..Im sorry ive been so troublesome, thank you so much!

9. Oct 2, 2011

issacnewton

10. Oct 2, 2011

issacnewton

mta, i think you should REALLY know how to complete squares BEFORE you take on calculus. have you had pre-calculus ?